Solution
Part B
Investigate the water-gas shift at AM1 and B3LYP/6-31G(d,p) level. Calculate the molar reaction energy, reaction enthalpy and free molar reaction enthalpy (Gibbs energy) at 0K and 298.15K for the conversion process:
B3LYP/6-31G(d,p)
| E /a.u. | E(zpe) /a.u. | E(298) /a.u. | H(298) /a.u. | G(298) /a.u. | |
| CO | -113,309454 | -113,304423 | -113,302062 | -113,301118 | -113,323561 |
| H2O | -76,419737 | -76,398373 | -76,395538 | -76,394594 | -76,416030 |
| CO2 | -188,580937 | -188.569348 | -188.566700 | -188.565756 | -188.590067 |
| H2 | -1,178539 | -1,168369 | -1,166008 | -1,165064 | -1,179856 |
| Reaction | DE /kcal/mol | DE(zpe) /kcal/mol | DE(298) /kcal/mol | DH(298) /kcal/mol | DG(298) /kcal/mol |
| -19,00 | -21.91 | -22,03 | -22,03 | -19.03 | |
| experimental value | -9,84 |
| AM1 | H(298) /a.u. | H(298) /kcal/mol |
| CO | -0,009086 | -5,70030544 |
| H2O | -0,0944287 | -59,24456638 |
| CO2 | -0,127281 | -79,8560994 |
| H2 | -0,0082606 | -5,18270044 |
| reaction | DH(298) /a.u. | DH(298) /kcal/mol |
| -0,032027 | -20,10 |
Part B
Discuss the difference between the two methods and between the different energies (enthalpies). How would you characterize the influence of the vibrational temperature correction? Compare the estimated values with the experimentally determined heat of formation. Calculate the equillibrium constant at 298°C.
Discussion
DE(298) = DH(298) since the work term is zero (no volume work). The main correction is the zero point vibrational energy correction which has the largest influence on the reaction energy. The thermal vibrational correction can be neglected. Moreover, the entropy has a huge influence on the equillibrium. Comparison of the experimentally determined heat of formation with the calculated value shows a fairly bad agreement. Both methods (AM1 and B3LYP) give a similar result.
DRG(298) = -RT lnK
K = exp[-DRG(298)/RT] = 8.78 .1013